Question

fosfat de magenziu , formula, felul si nr atomilor, r.m , M , c.p % , calculati cantitatea de O existenta in 200g subs

va rog, dau coronița!!​

Answer 1

Răspuns:

Formula :Mg3(Po4)2

Mg:P:O=3:2:8

Mg:P:O=3×AMg:2×AP:8×AO

=3×24:2×31÷8×16

=72:62:128

=36:31:64

%Mg=?

%P=?

%O=?

MMg3(Po4)2=3×AMg+2×AP+8×AO=

=72+62+128=

=134+128=

=262

1molMg3(Po4)2=262 g

262g Mg3(Po4)2.... 72g Mg.....62g P......128g O

100g Mg3(Po4)2------x---------------y------------z

x=100×72\(supra) 262=27,48% Mg

y=100×62\262=23,66% P

z=100×128\262=48,85% O

100g Mg3(Po4)2..... 128g O

200g Mg3(Po4)2.........x

x= 200×128/(supra) 100 =256g